prove that 3-√2 is irrational
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Hey mate!
Here is yr answer...
Let us assume 3-√2 is irrational!
let, 3-√2 = a/b (a, b € Z)
3-√2 = a/b
3 - a/b = √2
√2 = 3-a/b
√2 = 3b-a/b
For any two integers a, b RHS (3b-a/b) is rational!
Here, LHS (√2) is irrational!
A rational and irrational are never equal!
Therefore, 3-√2 is irrational
Hope it helps u..
Here is yr answer...
Let us assume 3-√2 is irrational!
let, 3-√2 = a/b (a, b € Z)
3-√2 = a/b
3 - a/b = √2
√2 = 3-a/b
√2 = 3b-a/b
For any two integers a, b RHS (3b-a/b) is rational!
Here, LHS (√2) is irrational!
A rational and irrational are never equal!
Therefore, 3-√2 is irrational
Hope it helps u..
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